Problem — Congruence with the altitudes

Statement

Given the triangle $ABC$ with $AC\cong BC$ and the altitudes $AH\perp CB$ and $BK\perp CA$, prove that $\widehat{ABK}\cong\widehat{BAH}$.

The isosceles triangle $ABC$ with the altitudes $AH$ and $BK$ relative to the oblique sides.

Solution

By the second congruence criterion, $\triangle ABK \cong \triangle BAH$ because:

• $AB$ is a common side;
• $\widehat{AKB} \cong \widehat{BHA} = 90^\circ$ (altitudes);
• $\widehat{KAB} \cong \widehat{HBA}$ (base angles of the isosceles triangle, since $AC\cong BC$).

From the congruence of the triangles it follows that $\boxed{\widehat{ABK}\cong\widehat{BAH}}$. $\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Altitude · Angle · Congruence criteria · Isosceles triangle
Skills: Proving · Synthetic geometry
Exercise type: Proof · Geometric problem