Question

ABC is right triangle right angled at B. prove that the perpendicular from B divides the triangle into two Triangles are similar each other and to original Triangle​

Answer 1

Refer the attachment for your answer

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Answer 2

$\bigtriangleup ADB \sim \bigtriangleup BDC \sim \bigtriangleup ABC$

Step-by-step explanation:

given : in triangle ABC ,

$\angle ABC = 90^\circ$

BD ⊥ AC

to prove :

$\bigtriangleup ADB \sim \bigtriangleup ABC \\\\\bigtriangleup BDC \sim \bigtriangleup ABC \\\\\bigtriangleup ADB \sim \bigtriangleup BDC$

Proof : in triangle ADB and ABC

$\angle DAB \cong \angle BAC ..(common)\\\angle ADB \cong \angle ABC= 90^\circ \\\bigtriangleup ADB \sim \bigtriangleup ABC (by AA)..(1)$

Similarly

in triangle BDC and ACB

$\angle BCD \cong \angle ACB ..(common)\\\angle BDC \cong \angle ABC= 90^\circ \\\bigtriangleup BDC \sim \bigtriangleup ABC (by AA)..(2)$

from 1 and 2

$\bigtriangleup ADB \sim \bigtriangleup BDC ..(3)$

From 1 ,2 and 3 we get

$\bigtriangleup ADB \sim \bigtriangleup BDC \sim \bigtriangleup ABC$

Hence proved

#Learn more:

AM is a median of a triangle ABC

Is AB + BC+CA>2 AM? (Consider the sides of triangle  triangle ABM and triangle  AMC.)​

https://brainly.in/question/13122021