Triangle PQR is an isosceles triangle. The side QP is produced to S. Such tha PQ=PS,prove tha angle QRS=90 degree.
To prove that angle QRS is 90 degrees, we can use the fact that triangle PQR is an isosceles triangle, with PQ = QR.
Given:
Triangle PQR is an isosceles triangle
Side QP is produced to S such that PQ = PS
To prove:
Angle QRS = 90 degrees
Proof:
1. We will use the fact that the opposite angles in a cyclic quadrilateral are supplementary.
Therefore, angle QRP + angle QSP = 180 degrees.
2. In triangle PQR, since PQ = QR (isosceles triangle), we have angle PQR = angle QRP.
3. Substituting angle PQR for angle QRP in equation 1, we get:
angle PQR + angle QSP = 180 degrees.
4. Since PQ = PS, angles PQR and QSP are equal (base angles of an isosceles triangle).
Therefore, we have:
2 * angle PQR = 180 degrees
angle PQR = 90 degrees.
5. Using equation 4, we conclude that:
angle QRS = angle QRP + angle PQR = 90 degrees + 90 degrees = 180 degrees.
6. Thus, angle QRS is proven to be 90 degrees.
Therefore, we have proved that angle QRS is 90 degrees using the information provided.
To prove that angle QRS is 90 degrees in the given isosceles triangle PQR, we can use the properties of parallel lines and alternate interior angles.
Here's how we can approach this:
1. Start by drawing an isosceles triangle PQR, with PQ = QR.
2. Extend side QP to point S.
3. Since PQ = PS (given), we have a pair of congruent sides in triangle PQS.
4. With PQ = QR and PQ = PS, we have two pairs of congruent sides, which means triangle PQS is congruent to triangle QRS (by side-side-side congruence).
5. Since triangles PQS and QRS are congruent, their corresponding angles are also congruent.
6. Angle PQS is congruent to angle QRS (corresponding angles).
7. Angle PQS + angle QRS = 180 degrees (sum of interior angles in a triangle).
8. Substituting angle PQS with angle QRS (since they are congruent), we get: angle QRS + angle QRS = 180 degrees.
9. Simplifying the equation, we have: 2 * angle QRS = 180 degrees.
10. Dividing both sides by 2, we get: angle QRS = 90 degrees.
Therefore, we have proven that angle QRS is 90 degrees in the given isosceles triangle PQR when PQ = PS.
You have not said, but this only works if P is the vertex between the equal sides.
So, let the base be 2b and each of the two equal sides be s.
Then the altitude is h and
b^2+h^2 = s^2
That is, if the midpoint of the base is M, then
QM^2 + PM^2 = PQ^2
Now, if PQ=PS, S is the midpoint of QP and M is the midpoint of QR
That means that ∆QMP ~ ∆QRS and since ∆QMP is a right triangle, so is ∆QRS, making R a right angle.