Answer:
QC - PB = 6
Step-by-step explanation:
Given PQ is parallel to BC then it divides the sides proportionally, that is
=
=
=
, then
=
( cross- multiply )
8PB = 5QC → *
Given PB + QC = 26 , then
PB = 26 - QC
Substitute into *
8(26 - QC) = 5QC ← distribute left side
208 - 8QC = 5QC ( subtract 5QC from both sides )
208 - 13QC = 0 ( subtract 208 from both sides )
- 13QC = - 208 ( divide both sides by - 13 )
QC = 16
Thus PB = 26 - QC = 26 - 16 = 10
QC - PB = 16 - 10 = 6
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Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
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Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
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Here's what we have so far for quadrilateral DAEC
Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.
