Question

In triangle ABC,<BAC=90°, Seg AP is perpendicular to side BC such that B-P-C . D is the midpoint of side BC , then prove that 2AD²=BD²+CD²​

Answer 1

Answer:

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Answer 2

According the question:-

• ABC is a Δ.
• ∠ BAC = 90°
• AP is perpendicular to side BC.
• D is the midpoint of side BC .

We have to prove  that 2AD²=BD²+CD²​.

In Δ ABC:-

AB² + AC² = BC²

∴ AB² + AB² =( BD + DC)²               ( AB = AC ) given

2AB² ≠ ( BD + CD)²     ...  ( 1).

R.H.S:-

( BD)² + ( CD)²= ( BP + PD)² +( CP - PD)²

                      = (BP)² + 2 BP. PD + ( PD)² + ( CP)² - 2 CP. PD - (PD)²

In Δ ABP:-

( BP)²= ( AB)² - ( AP)²...    ( 2).

In Δ ACP:-

( PC)² = (AC)²- ( AP)²

          = ( AB)² - ( AP)²  ...  (3)

∴( BP)² = ( PC)²

BP= PC ... (4)

From R.H.S we get:-

(BP)² + 2 BP. PD + ( PD)² + ( CP)² - 2 CP. PD - (PD)²

= 2( BP² + PD²)  ...(5)

In Δ ABC  and ΔABP

• ∠A + 90°
• ∠B = X ( Let consider )
• ∠C =  (90°-X)

∴ Δ ABC  ≅ ΔABP

Hence :- PA/PB = AB/AC= AB/AB

∴ PA = PB

2( PA)² = 2(AD)²

∴2 AD² = BD² + CD²

Ans :- L.H.S = R.H.S

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