Are triangles ΔABD and ΔADC similar?
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Are triangles ΔABD and ΔADC similar?
Since side AB is equal to side AC, the triangle is isosceles.
AD divides BC into two equal parts, therefore BD = DC.
Also, AD is a common side.
From this, it follows that the triangles are similar according to the S.S.S (side side side) criterion.
Yes, according to S.S.S.
The height of 7 units shows that AD is perpendicular to BC, which is a key property of isosceles triangles. This confirms that AD bisects the base BC into equal segments.
In an isosceles triangle, when you draw a height from the apex (vertex A) to the base BC, it always bisects the base into two equal segments. This is a fundamental property!
Congruent triangles are identical in size and shape (SSS proves congruence). Similar triangles have the same shape but may differ in size. Since these triangles are congruent, they're automatically similar too!
The 15 units represent the equal sides AB and AC of the isosceles triangle. Since both triangles share these same sides, it actually supports the SSS similarity criterion.
You could also use SAS (Side-Angle-Side) since both triangles share the right angle at D, and have AD common plus either AB = AC. However, SSS is most direct here.
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