In triangle ABC, AB = AC. The altitude BH is drawn perpendicular to AC at H, and the altitude CK is drawn perpendicular to AB at K. Prove that BK = CH, and that KH is parallel to BC.

Step-by-step solution

1. 1

   Set up what the hypothesis gives you. Since $AB=AC$, triangle $ABC$ is isosceles at $A$, so the base angles are equal: $\widehat{ABC}=\widehat{ACB}$. The two altitudes create right angles $\widehat{BHA}=\widehat{BHC}=90^\circ$ and $\widehat{CKA}=\widehat{CKB}=90^\circ$. The whole proof runs on the symmetry of the figure about the axis through $A$.

2. 2

   Prove the two right triangles BKC and CHB are congruent. Compare $\triangle BKC$ and $\triangle CHB$:

   • $\widehat{BKC}=\widehat{CHB}=90^\circ$ (the altitudes),
   • $BC$ is a common hypotenuse,
   • $\widehat{KBC}=\widehat{HCB}$ (the equal base angles of the isosceles triangle).

   By hypotenuse-and-acute-angle, $\triangle BKC\cong\triangle CHB$.

3. 3

   Read off part (a). Corresponding legs of congruent triangles are equal, and $BK$ in the first corresponds to $CH$ in the second:

   $BK=CH$

   This is part (a). As a by-product, $CK=BH$ — the two altitudes to the equal sides have equal length.

4. 4

   Convert the equality into a ratio on the two sides. Since $AB=AC$ and $BK=CH$, subtracting gives $AK=AB-BK=AC-CH=AH$. Therefore

   $\frac{AK}{AB}=\frac{AH}{AC}$

   both sides being the same number. This is precisely the proportion the converse of Thales' theorem needs.

5. 5

   Apply the converse of Thales to get part (b). Points $K$ and $H$ lie on sides $AB$ and $AC$ of triangle $ABC$ and cut them in the same ratio, so the line $KH$ is parallel to the third side:

   $KH\parallel BC$

   Equivalently, $\triangle AKH\sim\triangle ABC$ by SAS similarity (common angle at $A$, proportional adjacent sides), which forces the corresponding angles $\widehat{AKH}=\widehat{ABC}$ and hence the parallelism.

6. 6

   Verify with coordinates. Placing $B=(-b,0)$, $C=(b,0)$, $A=(0,h)$ makes $AB=AC$ automatic. Computing $H$ and $K$ as perpendicular feet for six random pairs $(b,h)$ gives $BK=CH$ every time to within $10^{-9}$, $AK=AH$ likewise, and $K$ and $H$ always share the same $y$-coordinate — so $KH$ is horizontal, exactly like $BC$ ✓.