Pls solve this i will mark the right answer as brainlist
Answers
Answer:
In a triangle ABC, angle BAC = 90 degree and AD is perpendicular to BC then, \mathrm{BD} \times \mathrm{CD}=\mathrm{AD}^{2}BD×CD=AD
2
Solution:
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
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