Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal
Jump to navigation
Jump to search
Theorem
Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$.
Then $CE = CF$.
Proof
$\triangle CDE$ and $\triangle CDF$ are right triangle where $CE$ and $CF$ are the hypotenuses.
We have:
- $\angle CDE = \angle CDF$ as both are right angles.
- $\angle DCE = \angle DCF$ by hypothesis.
- $CD$ is common.
Thus by Triangle Angle-Side-Angle Congruence, $\triangle CDE$ and $\triangle CDF$ are congruent.
Hence the result.
$\blacksquare$
Sources
- 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.21 \ \text{(ii)}$