Question

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Answer 1

Step-by-step explanation:

The figure is attached below

In Triangle ABC, angle BAC = 90 degree and AD is perpendicular to BC

Given in the question a perpendicular is drawn AD on BC from angle BAC which is 90 degree

So, these two triangles ABD and CAD are similar

\Delta \mathrm{ABD} \approx \Delta \mathrm{CAD}ΔABD≈ΔCAD

By Corresponding part of Similar Triangles (CPST),

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

This proves that the ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

\text {we can write } \frac{B D}{A D}=\frac{A D}{C D}we can write

AD

BD

=

CD

AD

By cross-multiplication we get,

\mathrm{BD} \times \mathrm{CD}=\mathrm{AD} \times \mathrm{AD}=\mathrm{AD}^{2}BD×CD=AD×AD=AD

2

\mathrm{BD} \times \mathrm{CD}=\mathrm{AD}^{2}BD×CD=AD

2

Which is our Required Expression

Hence , option 2 is correct

Learn more about triangles

In triangle ABC, angle BAC=90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DEx(AB+AC)=ABxAC.

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