In the triangle ABC, ab = AC, CD is the height on the edge AB, prove 2 ∠ BCD = ∠ a

In the triangle ABC, ab = AC, CD is the height on the edge AB, prove 2 ∠ BCD = ∠ a

Make AE vertical BC and CD intersect F through point a
Because AB = AC, the angle BAF = the angle caf
Because angle ADF = angle FEC = 90 degree angle AFD = angle CFE
So DAF = BCD
And because the angle BAF = the angle CAF
SO 2 ∠ BCD = ∠ a

As shown in the figure, in the triangle ABC, CD is perpendicular to ab at point D, AC square = ad × AB, and the proof is: ∠ ACB = 90 degree

From CD perpendicular to ab at point d
∠adc= 90°
From the square of AC = ad × ab
ac/ab=ad/ac
∠a= ∠a
The triangle ABC is similar to the triangle ACD
That is, ADC = ∠ ACB = 90 degree
1 | comments
Perfect my answer

In the triangle ABC, D is a point on the edge of AB, and the square of AC = ad times AB, it is proved that CD is perpendicular to ab
We're missing BC squared = BD times ab

Certification:
∵AC²=AD*AB
∴AC/AB=AD/AC
∵∠A=∠A
∴△ACD∽△ABC
∴∠ACD=∠B
∵BC²=BD*AB
Similarly, we can get ∠ B = ∠ ACD
∵∠A+∠B+∠ACD+∠BCD=180°
∴∠ACD+∠A=90°
∴∠ADC=90°
That is CD ⊥ ab

As shown in the figure, D is a point on AB, the square of AC = ab × ad, the square of BC = BD × ab. question: is △ ABC a right triangle? Please explain the reason

It is proved that: ∵ AC & # 178; = ab × ad ∵ AC / ad = AB / AC ∵ CAD = basc ∵ ABC ∵ ACD ∵ ACD = B ∵ BC ∵ 178; = BD × ab ∵ BC / BD = AB / BC ∵ CBD = ABC ∵ ABC ∵ CBD ∵ BCD = a ∵ ACB = ACD + BCD = a + B