 UPSKILL MATH PLUS

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right angles
corresponding pair of sides
opposite side
$$CY$$ is perpendicular to $$AB$$
$$\triangle ABX$$
$$AC$$
$$BX$$ and $$CY$$
$$\triangle ACY \cong \triangle ABX$$
$$ACY$$ Prove that the altitudes are $$BX$$ and $$CY$$ are equal if triangle $$ABC$$ is isosceles with $$AB = AC$$.

Proof:

It is given that
are altitudes of triangle $$ABC$$.

An altitude is a perpendicular line segment drawn through the vertex of the triangle to the
.

Here, $$CY$$ is an altitude of $$AB$$, and $$BX$$ is an altitude of
.

Hence,
and $$BX$$ is perpendicular to $$AC$$.

To prove that the altitudes are equal, let us consider $$\triangle ACY$$ and
.

Here, $$AB = AC$$ [Given]

Also, $$\angle AXB = \angle AYC$$, as the altitudes meet the sides at
.

Also, $$\angle A$$ is common to both triangles
and $$ABX$$.

Here, two corresponding pairs of angles and one
are equal.

Thus by  congruence criterion, $$\triangle ACY \cong \triangle ABX$$.

Since
and by CPCT, the altitudes $$CY$$ and $$BX$$ are equal.