Question

In ∆ABC, from A and B altitudes AD and BE are drawn, such that AE is perpendicular to AC and AD is perpendicular to BC. Prove that ∆ADC ~ ∆BEC. Is ∆ADB ~ ∆ AEB and ∆ADB ~ ∆ADC?

Answer 1

Answer:

Step-by-step explanation:

The given question is as follows:-

In∆ABC,AD is perpendicular on BC and BE is perpendicular on AC . AD and

BE intersect each other at F. If BF= AC , then what is the value of angle ABC ?

Solution:-

In right angled ∆ BEC let angle BCE=C ,then angle CBE(or DBF)=90°-C……(1)

In right angled ∆ ADC ,angle DCA=C ,then angle DAC=90-C…………..(2)

In right angled ∆ FDB , cosDBF=BD/BF. or cos(90-C)= BD/BF………..(3)

In right angled ∆ ADC ,cos DAC = AD/AC. or cos (90-C)= AD/AC……….(4)

From eqn. (3) and (4)

BD/BF=AD/AC. Putting BF=AC (given).

BD/AC=AD/

(1) Angle FDB = angle CDA … ..(each right angle )

(2) side BF = side AC…..(given )

(3) Angle DBF = angle DAC =90-C…..(proved above)

Thus ,∆FDB is congruent ∆ CDA

Hence side BD = side AD…………(3)

Let angle ABC=x°

In right angled ∆ADB , side BD= side AD (proved above)

Thus , angle BAD = anglen ABD = x°

x+x+90°=180°

or. 2x=90°. => x = 90°/2= 45°. Answer.