A triangle $A B C$ is right-angled at $A$ . $A M$ is drawn perpendicular to $B C$ . Prove that $∠ BAM = ∠ ACB$ .

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Solution

Applying the similarity of triangles:
In $∆ B A C and ∆ B M A$ ,
$∠ B = ∠ B (Commonangle) ∠ B A C = ∠ B M A = 90 °$
So, by Angle-Angle (AA) similarity, $∆ B A C ~ ∆ B M A$ .
Therefore,
$∠ B C A = ∠ B A M \Rightarrow ∠ A C B = ∠ B A M$
Hence, $∠ BAM = ∠ ACB$ is proved.

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A triangle ABC is right-angled at A. AM is drawn perpendicular to BC. Prove that ∠BAM=∠ACB.

Applying the similarity of triangles:In ∆BACand∆BMA,∠B=∠B(Commonangle)∠BAC=∠BMA=90°So, by Angle-Angle (AA) similarity, ∆BAC~∆BMA.Therefore,∠BCA=∠BAM⇒∠ACB=∠BAMHence, ∠BAM=∠ACB is proved.

A triangle $A B C$ is right-angled at $A$ . $A M$ is drawn perpendicular to $B C$ . Prove that $∠ BAM = ∠ ACB$ .

Applying the similarity of triangles:
In $∆ B A C and ∆ B M A$ ,
$∠ B = ∠ B (Commonangle) ∠ B A C = ∠ B M A = 90 °$
So, by Angle-Angle (AA) similarity, $∆ B A C ~ ∆ B M A$ .
Therefore,
$∠ B C A = ∠ B A M \Rightarrow ∠ A C B = ∠ B A M$
Hence, $∠ BAM = ∠ ACB$ is proved.