Question

show that the bisector of rectangle of an isosceles triangle the base at right angle​

Answer 1

Since we want to prove that the bisector of the vertical angles of an isosceles triangle bisects the base at right angle, we need to consider the triangle congruence.

There is a theorem which says that if two angles, a and b, are equal, then they have two opposite sides, r and t, respectively, which are also equal.

Then we have an isosceles triangle, and the bisector of the vertical angles will divide the angle ∠ABC in two smallest equal angles.

Then if AB = AC we can confirm that this bisects the base at the middle point of BC, and makes a right angle.

$\underline\green{\:\textsf{Step-by-step Explanation}}$

Isosceles triangle is shown in the figure above.

∠B = ∠C

Sides AB = AC.

The AD line is a angular bisect.

=> ∠BAD = ∠DAC = ∠A/2

In ΔABC, ∠A + ∠B + ∠C = 180

Since ∠B = ∠C,

∠A + 2∠B = 180

In ΔBAD, ∠B + ∠ADB + ∠DAB = 180

Since ∠DAB = ∠A/2

∠B + ∠ADB + ∠A/2 = 180

2∠B + 2∠ADB + ∠A = 360 (Multiply both sides with 2)

From above ∠A + 2∠B = 180

Hence 180 + 2∠ADB = 360

∠ADB = 90.

Similar way, we can prove that ∠ADC = 90.

Now in ΔADB, ΔADC,

AB = BC, AD = AD and all angles are same.

That means ΔADB and ΔADC are congruent.

Therefore BD = CD.

That implies, D is center of BC. AD bisects BC.

Answer 2

Answer is in the attachment.

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