The correct answer is C. <span>Similar polygons are never congruent.</span>
Answer:
See explanation
Step-by-step explanation:
In ΔABC, m∠B = m∠C.
BH is angle B bisector, then by definition of angle bisector
∠CBH ≅ ∠HBK
m∠CBH = m∠HBK = 1/2m∠B
CK is angle C bisector, then by definition of angle bisector
∠BCK ≅ ∠KCH
m∠BCK = m∠KCH = 1/2m∠C
Since m∠B = m∠C, then
m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH (*)
Consider triangles CBH and BCK. In these triangles,
- ∠CBH ≅ ∠BCK (from equality (*));
- ∠HCB ≅ ∠KBC, because m∠B = m∠C;
- BC ≅CB by reflexive property.
So, triangles CBH and BCK are congruent by ASA postulate.
Congruent triangles have congruent corresponding sides, hence
BH ≅ CK.
Answer:an isosceles triangle with PQ=QR (option C)
Step-by-step explanation:
The perimeter of Jake window is 16x - 2 inches
<h3><u>Solution:</u></h3>
Given that height of Jake window is 5x - 3 inches
Also given that width is 3x + 2 inches
To find: perimeter of Jake window
We know that, in general a window is of rectangular shape
<em><u>The perimeter of rectangle is given as:</u></em>
Perimeter of rectangle = 2(length + width)
Substituting the values we get,
Perimeter of rectangle = 2(5x - 3 + 3x + 2)
Perimeter of rectangle = 2(8x - 1) = 16x - 2
Hence, the perimeter of the window is 16x – 2 inches