In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If $AD \perp BC$ and $EF \perp AC$ , prove that $\triangle ABD \sim \triangle ECF$ .

Answer: $\triangle ABD \sim \triangle ECF$

Step-by-step solution

Step 1: Identify given information and properties of isosceles triangle

We are given that triangle ABC is an isosceles triangle with sides AB equal to AC. In an isosceles triangle, the angles opposite to the equal sides are also equal. Therefore, angle B is equal to angle C.

Step 2: Identify right angles from perpendicular lines

We are given that AD is perpendicular to BC, which means that angle ADB is a right angle, measuring 90 degrees. Similarly, EF is perpendicular to AC, which means that angle EFC is also a right angle, measuring 90 degrees.

Step 3: Apply AA similarity criterion

Now we compare triangle ABD and triangle ECF. From step 1, we know that angle B is equal to angle C. From step 2, we know that angle ADB and angle EFC are both 90 degrees. Since two angles of triangle ABD are equal to two corresponding angles of triangle ECF, by the Angle-Angle (AA) similarity criterion, the two triangles are similar.

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In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD⊥BCAD \perp BCAD⊥BC and EF⊥ACEF \perp ACEF⊥AC, prove that △ABD∼△ECF\triangle ABD \sim \triangle ECF△ABD∼△ECF.