Question

in figure angle BAC equal to 90 degree, AD is the bisector. If PE perpendicular AC prove that DE into (AB + AC) equal to AB into AC​

Answer 1

Answer:

you can use the formula of cpct

Answer 2

DE ×(AB + AC) = AB×AC

Step-by-step explanation:

given :  $\angle BAC = 90^\circ$ and AD is the bisector

to prove :  DE ×(AB + AC) = AB×AC

proof :

In triangle ABC

$\angle BAC = 90^\circ$

let us draw DF ⊥ AB

then in triangle ADE and ADF

AD = AD (common)

$\angle E = \angle F = 90^\circ$

$\angle DAF =\angle DAE$  

Therefore ,

$\bigtriangleup ADE \cong \bigtriangleup ADF$

By AAS congruency then

DE = DF (by cpct )

area of triangle ABD and ACD

area of ABD

$\frac{1}{2} \times AB \times DF \\\frac{1}{2} \times AB \times DE$(DE = DF)

area of ACD = $\frac{1}{2} \times AC \times DE$

Then

area of triangle ABC = area of ABD +area of ACD

area of triangle ABC = $\frac{1}{2} \times AC \times DE+ \frac{1}{2} \times AC \times DE$

area of triangle ABC = $\frac{1}{2} \times DE(AB +AC)$

Area of ABC = $\frac{1}{2} \times AB \times AC$

Then ,

$\frac{1}{2} \times DE(AB +AC)$= $\frac{1}{2} \times AB \times AC$

DE ×(AB + AC) = AB×AC

hence proved

#Learn more:

In the figure angle BAC is equal to 90 degree and AD perpendicular to BC then prove that BD X CD is equal to AD square​

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