Simpler way. Angle ABQ = X (ABC is isosceles). Draw BP. Triangles ABP & QBP are mirror congruent (side-side-side),so angle PQB = X. This is exterior angle of isosceles triangle PQC, so angle PCQ = X/2. In ABC, sum of angles = 180 degrees. X + X + X/2 = 180. 5X = 360. X = 360/5 = 72 degrees.
I assigned AP=PQ=QC=1 and AB=BQ=PC="a". Since the triangle is isosceles ∠ABQ = x and it can be shown that ∠APQ = 2x. Using the Law of Cosines in two ways to find AQ results in cos(x) = (a² - 2)/2. This means (a² - 2)/2 = (a/2)/(a + 1) which results in a = (1 + √5)/2. So cos(x) = (√5 - 1)/4 from which x = 72º.
If ∠C=α , then ∠PQB =2α ∆PBQ is Congruent to ∆PBA by SSS, hence ∠A=∠PQB Since ∆ABC is Isosceles, ∠A =∠B Summing up angles in ∆ABC, 5α= 180 X = ∠A = 2α= 72°
How do you figure this equivalence below? If ∠C=α , then ∠PQB =2α EDIT1 Is it by considering that they can be seen as angle at the circumference and angle at the center insisting on the same arc of a circle of radius QC=QP? EDIT2 Oh now I see it's much simpler than that, because the angle in C = CPQ = a then PQC must be 180-2a (the sum of internal angles of a triangle being 180) and then PQB is the complement to 180 of PQC so it must be 180 - (180 -2a) = 2a Very good.
I feel that I should also know the ratio of AP/AB (Working it out at the end: sine X = (1/2) PC/PQ) AP/AB =1/(2sine(X) ) In triangle ABC , angle B equals X , because AC = AB. Angle C = 180 - X - X In triangle CPQ , angle P = 180 - X -X and angle Q = 4X In kite APQB A = X and we have two expressions for Q . It is X because triangles APB, QPB are (side,side,side) congruent. It is 360-4X (the sum of QCP and QPC) 5X = 360 360/5 = 72 X=72 degrees Answer Check: 72 +72 + 36 = 180 36+36+ 108 = 180 72 + 144+72 +72 = 360 degrees in the vertices of APQB.
AC=BC ; PQ=QC---> ABC y PQC son triángulos isósceles---> Aº=Bº--->Cº=180-2X =QPCº---> PQCº=180-2(180-2X)=4X-180---> BQP=180-4X+180=360-4X ---> Por simetría respecto al eje BP, Aº=BQPº---> X=360-4X---> X=72º. Gracias y un saludo cordial.
Simpler way. Angle ABQ = X (ABC is isosceles). Draw BP. Triangles ABP & QBP are mirror congruent (side-side-side),so angle PQB = X. This is exterior angle of isosceles triangle PQC, so angle PCQ = X/2. In ABC, sum of angles = 180 degrees. X + X + X/2 = 180. 5X = 360.
X = 360/5 = 72 degrees.
i did the same thing as you
I assigned AP=PQ=QC=1 and AB=BQ=PC="a". Since the triangle is isosceles ∠ABQ = x and it can be shown that ∠APQ = 2x. Using the Law of Cosines in two ways to find AQ results in cos(x) = (a² - 2)/2. This means (a² - 2)/2 = (a/2)/(a + 1) which results in a = (1 + √5)/2. So cos(x) = (√5 - 1)/4 from which x = 72º.
If ∠C=α , then ∠PQB =2α
∆PBQ is Congruent to ∆PBA by SSS, hence ∠A=∠PQB
Since ∆ABC is Isosceles, ∠A =∠B
Summing up angles in ∆ABC,
5α= 180
X = ∠A = 2α= 72°
Excellent! For ∆PBQ and ∆PBA, you need to construct PB, but I'm sure viewers will figure that out!
How do you figure this equivalence below?
If ∠C=α , then ∠PQB =2α
EDIT1
Is it by considering that they can be seen as angle at the circumference and angle at the center insisting on the same arc of a circle of radius QC=QP?
EDIT2
Oh now I see it's much simpler than that, because the angle in C = CPQ = a then PQC must be 180-2a (the sum of internal angles of a triangle being 180) and then PQB is the complement to 180 of PQC so it must be 180 - (180 -2a) = 2a Very good.
@@3r4kl3s
Exterior angle of Isosceles ∆PQC.
@@3r4kl3s At 3:00 in the video, PreMath has shown that
Angle ACB = "α" ; Angle BQP = "x"
x = 2α --> α=½x
Isosceles triangle ABC:
2x+α = 2x+½x= 180°
x = 180°/2,5 = 72° (Solved √)
Let PC=a, PA=b, and
I started down that route - and gave up!
We think Alike : )
Gracias for explaining this point-of-view in detail-Top Solution !
I feel that I should also know the ratio of AP/AB (Working it out at the end: sine X = (1/2) PC/PQ) AP/AB =1/(2sine(X) )
In triangle ABC , angle B equals X , because AC = AB. Angle C = 180 - X - X
In triangle CPQ , angle P = 180 - X -X and angle Q = 4X
In kite APQB A = X and we have two expressions for Q . It is X because triangles APB, QPB are (side,side,side) congruent. It is 360-4X (the sum of QCP and QPC)
5X = 360 360/5 = 72 X=72 degrees Answer
Check: 72 +72 + 36 = 180 36+36+ 108 = 180 72 + 144+72 +72 = 360 degrees in the vertices of APQB.
AC=BC ; PQ=QC---> ABC y PQC son triángulos isósceles---> Aº=Bº--->Cº=180-2X =QPCº---> PQCº=180-2(180-2X)=4X-180---> BQP=180-4X+180=360-4X ---> Por simetría respecto al eje BP, Aº=BQPº---> X=360-4X---> X=72º.
Gracias y un saludo cordial.
Cool ! The famous golden triangle again.
二等辺三角形の底角∠Cと対をなすもう一つの底角の対頂角が、
QPとBAの延長線の交点を頂点とする二等辺三角形の頂角と対をなす二等辺三角形の底角になる.
二等辺三角形の底角の和∠BQPが、
二等辺三角形の一つの底角と等しく、
内角の和が180°
つまり頂角θの2倍の底角2θが2つあり、
θ+2θ+2θ=180°
θ=36°
X=2θ=72°
Из дельтоида ABQP 5х=360⇔360/5=72. Также гипотенуза ВР делит его на равносторонние ▲АВР и ▲BQP, подобные друг другу и общему ▲АВС.
С какого перепоя:
1. Сумма углов ABQP равна 5х ?!
2. ВР - гипотенуза ?!
3. ΔABP и ΔBQP - равносторонние ?!
4. Они подобны друг другу и ΔABC ?!
APQ = 2 . ABQ A=B=Q A+B+Q+P =5X 360/5 = X=72
X=72
{60°A+60°B+60°C}=180°ABC 60^60^60ABC 2^30^2^30^2^30 1^10^1^10^2^3^10 2^3^5^5 2^3^2^3^2^3 1^1^1^1^2^3 2^3 (ABC ➖ 3ABC+2).
72 degrees
❤
After all, it seems to be an easy solve problem. But it isn't. Congrats.