Math Problem Statement
Triangle $ABC$ has $\angle ACB = 90^{\circ}$. Point $D$ is on side $\overline{AB}$ such that $AD = CD$. Prove that $D$ is the midpoint of $\overline{AB}$. [asy] pair A, B, C, D; A = (-5,0); D = (0,0); B = (5,0); C = (-3,4); draw(A--B--C--cycle); draw(C--D, dotted); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, S); [/asy] Since A D
C D , is isosceles, giving the equal angles
; let the measure of these angles be x ∘ . Then, ∠ D C B
−
90 ∘ − x ∘ . Since the angles of add to 180 ∘ , ∠ A B C + ∠ A C B + ∠ B A C
90 ∘ + x ∘
180 ∘ , and solving gives
90 ∘ − x ∘ . Therefore,
, so is isosceles. Therefore, B D
C D
A D , and we conclude that D is the midpoint of ¯¯¯¯¯¯¯¯ A B . ∠ A C B ∠ D A C ∠ D B C ∠ D C A ∠ D C B △ A D C △ C D B △ A B C ∠ C D A ∠ C D B
Solution
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Math Problem Analysis
Mathematical Concepts
Geometry
Isosceles Triangles
Right Triangles
Formulas
-
Theorems
Isosceles Triangle Theorem
Angle Sum Property of Triangles
Suitable Grade Level
Grades 9-10
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