A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Step-by-Step Solution
Key Concept: Tangents from external points are equal: $AP = AS, BP = BQ, CR = CQ, DR = DS$.
Stepwise Solution:
Let the circle touch sides $AB, BC, CD, DA$ at $P, Q, R, S$ respectively. [0.5 Mark]
Tangents from external points are equal:
$AP = AS$ -- (1)
$BP = BQ$ -- (2)
$CR = CQ$ -- (3)
$DR = DS$ -- (4). [1.0 Mark]
Adding (1) + (2) + (3) + (4):
$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \Rightarrow AB + CD = AD + BC$. Proved! [0.5 Mark]
Marking Scheme:
• Writing tangent equality equations for all 4 vertices: 1.0 Mark
• Adding equations and grouping terms to get $AB + CD = AD + BC$: 1.0 Mark
Correct Answer: