Question

let ABC be a triangle right angled at B and D be the midpoint of AC show that DA = DB = DC​

Answer 1

PA = PC = PB in the right triangle ABC

Step-by-step explanation:

Let P be the mid point of the Hypotenuse. of the right triangle ABC, right angled at B

Draw a line parallel to BC from P meeting at D 

Join PB

In triangles PAD and PBD

Angle PDA = angle PDB (90 each due to converse of mid point theorem)

PD = PD (common)

AD = DB (D is mid point of AB)

So triangles PAD and PBD are congruent by SAS rule.

PA = PB (C.P.C.T)

but PA = PC (As P is mid point)

So, PA = PC = PB

Answer 2

Hence, DA=DB=DC.

Step-by-step explanation:

Through D, draw DE||BC, meeting AB at E.

Now, ∠AED=∠ABC=90∘ [corresponding angle]

Therefore, ∠BED=∠AED=90∘[∵∠AED+∠BED=180∘].

Now, in △ABC, it is given that D is the midpoint of AC and DE||BC (by construction).

Therefore, E must be the midpoint of AB (by converse of midpoint theorem).

Therefore,AE=BE.

Now, in △ AED and BED, we have

• AE=BE (proved), ED=ED  (common),
• ∠AED=∠BED (each equal to 90∘).

Therefore,△AED≅△BED

Therefore,DA=DB.

But, DA=DC [∵ D is the midpoint of AC].

Hence, DA=DB=DC.