Geometry Problem 1623
Two Adjacent Squares and Bounding Square Proof
Problem Statement
Let $A, B, C$ be three collinear points in that order. Two squares, $S_1 = ABDE$ and $S_2 = BCFG$, are constructed on the same side of line $ABC$, sharing the vertical line through $B$ that contains sides $BD$ and $BG$.
Let $Q$ be the bounding rectangle circumscribing $S_1 \cup S_2$ with sides parallel and perpendicular to $AG$.
If $h$ is the length of segment $AG$ and $m$ is the altitude from $B$ to $AG$ in right triangle $ABG$:
To Prove:
$Q$ is a square with side length:
$$L = h + m$$Strategic Synthetic Hints
- Idea 1 — Primary Congruence: Compare the two main right triangles embedded across the two squares. Showing they are congruent immediately unlocks the angle relationship between their hypotenuses.
- Idea 2 — Corner Congruences: Look at the right triangles formed at the four outer corners of rectangle $Q$. Establishing their congruence to the primary triangle proves that all four sides of $Q$ are equal, confirming $Q$ is a square.
- Idea 3 — Central Altitude Segment: Trace how dropping the altitude in the main right triangle creates a key segment that translates via congruent shapes directly onto the outer boundary.
- Idea 4 — Side Partition: Combine the segments formed along any side of square $Q$ to see how $h$ and $m$ add up directly to equal $L$.
Foundation Theorems — Direct Tools
Pattern note: The perpendicular orientation of rectangle $Q$ inherently aligns with hypotenuse $CD \perp AG$ via rotational congruence. Projecting the vertices of $S_1 \cup S_2$ isolates a central square of side $m = ab/h$, establishing the clean linear identity $L = h + m$.
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