in any right angle triangle if a perpendicular is drawn from the right angle point on the hypothesis the two triangle on both and each of them is similar to the right angle
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If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.
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Hint: First we need to draw the diagram using the given condition in the question and we have to prove :
ΔADB∼ΔABC ΔBDC∼ΔABC Δ ADB∼ΔBDC
Using the properties of similar triangles.
Complete step by step answer:
We are given that ΔABC
is right angled at B and a perpendicular is drawn from B to hypotenuse AC at D.
So we need to prove that
ΔADB∼ΔABC ΔBDC∼ΔABC Δ ADB∼ΔBDC
Firstly we will consider ΔADB
and ΔABC
∠ADB=∠ABC
(90∘
angle)
∠A=∠A
(common angle)
Since two angles are equal therefore the third corresponding angles of the triangles will be equal
Therefore,
∠ABD=∠ACB
Hence ΔADB∼ΔABC
by AAA criterion. …………(1)
Now, we will consider ΔBDC
and ΔABC
∠BDC=∠ABC
(90∘
angle)
∠C=∠C
(common angle)
Since two angles are equal therefore the third corresponding angles of the triangles will be equal
Therefore,
∠CBD=∠CAB
Hence ΔBDC∼ΔABC
by AAA criterion. …………………(2)
Since from (1) and (2)
ΔADB∼ΔABC
and ΔBDC∼ΔABC
Therefore, Δ ADB∼ΔBDC
. Hence proved.
Note:
The similarity of triangles has several criterions:
AAA (angle angle angle) criterion in which the angles of the similar triangles are equal
SAS(side angle side) criterion in which two sides of similar triangles are in the same ratio and the angle between them is equal.
ASA(angle side angle) criterion in which two angles of the similar triangles are equal and angle between them is equal.
SSS(side side side) criterion in which the ratio of the sides of the similar triangles are equal.
Explanation:
4 picture
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