Problem 4 · A quadrilateral with two right angles
Chained Pythagorean theorem across three right triangles.
Integer answer, at most 4 digitsThe figure shows a quadrilateral $ABCD$ and a segment from vertex $C$ to side $AD$. The two marked segments have the same length, and there are two right angles, also marked in the figure. Side $AD$ is divided into two segments of lengths 4 cm and 6 cm. How long is side $AB$?
Reasoned solution
Key idea: three right triangles share sides; applying the Pythagorean theorem three times, the unknown $DC$ cancels by itself.
Let $P$ be the point of side $AD$ reached by the segment from $C$, with $AP = 4$ and $PD = 6$. The marked equal segments are $PC$ and $BC$, and the right angles are at $D$ and at $B$.
Triangle $PDC$, right-angled at $D$:
Triangle $ADC$, right-angled at $D$ (because $AD$ passes through $P$ and $D$):
Subtracting the two equalities: $AC^{2} - PC^{2} = 64$, without knowing $DC$.
Finally, triangle $ABC$ is right-angled at $B$ and $BC = PC$: