Question

if P is the midpoint of hypotenuse AC of a right angle triangle ABC prove that BD = half AC

Answer 1

Given:

   Right angle triangle ΔABC

where ∡ABC=90∘

BD

divides AC, i.e., AD=DC

From D

, draw ED and FD⊥ to AB and BC

respectively

In DEBF

, Because ∡BED = ∡BFD = ∡FBE = 90∘, ∠EDF=90∘. Therefore, DEBF is a rectangle. Hence, BE = DF, ED = BF

.

In Δs AED

and DFC

,

   ∡EAD = ∡FDC

(AB//FD and AC

cuts it — corresponding angles).

∡DEA = ∡CFD=90∘

(Construction)

AD=DC

   (Given)

So ΔAED = ΔDFC

. So AE=FD. Therefore, AE=BE

.

In Δs AED,BED

,

   AE=BE

DE

is common

included angles ∡DEA = ∡DEB=90∘

   (Construction)

So, ΔAED=ΔBED

.

Therefore, BD=AD=AC/2.