in the given figure angle BAc =90 degree , AD perpendicular to BC . prove AO sqaure = BD× DC
Answers
Step-by-step explanation:
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.
https://brainly.in/question/11464220
In a triangle ABC angle BAC = 90 degree and AD is drawn perp. to BC. Prove that AD2 = BD.CD. Prove it using Pythagoras theorem
Step-by-step explanation:
okkkkkkkkkkkkkkkkkkkkkkkkkkk
