Prove the following statement. Let ABC be any right triangle, the right angle at point C. The altitude drawn from C to the hypotenuse splits the triangle into two right triangles that are similar to each other and to the original triangle?

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Answer 1 · 2018-05-23T08:17:06

See Below.

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According to the Question,

$DeltaABC$ is a right triangle with $/_C = 90^@$ , and $CD$ is the altitude to the hypotenuse $AB$ .

Proof:

Let's Assume that $/_ABC = x^@$ .

So, $angleBAC = 90^@- x^@ = (90 - x)^@$

Now, $CD$ perpendicular $AB$ .

So, $angleBDC = angleADC = 90^@$ .

In $DeltaCBD$ ,

$angleBCD = 180^@ - angleBDC - angleCBD = 180^@ - 90^@ - x^@ = (90 -x)^@$

Similarly, $angleACD = x^@$ .

Now, In $DeltaBCD$ and $DeltaACD$ ,

$angle CBD = angle ACD$

and $angle BDC = angleADC$ .

So, by AA Criteria of Similarity, $DeltaBCD ~= DeltaACD$ .

Similarly, We can find, $DeltaBCD ~= DeltaABC$ .

From that, $DeltaACD ~= DeltaABC$ .

Hope this helps.