Question

#3

3. Below you will find a series of constructions. Prove the statements al. , & (c). Construction IX: Locate a point on AB that is not between A and B such that the ratio of AP to BP is equal to the ratio of two given lengths a and b. Step 1: Choose a point not on AB such that AC-a. Step 2: At draw a line parallel to AC and located on the same side of the line AB as such that BD b. Step 3: Draw the ray CD to intersect the line AB at P. (a) Prover This construction is referred to as dividing a line segment externally so that the ratio of the total distance is to the extended distance is the same as the ratio of two given segments. Definition: A tangent to given circles that do not intersect, but the tangent intersects the line segment determine by the two centers of the circle, is called an internal tangent to the two circles. A tangent to two given circles that does not intersect the segment determined by the centers of the two given circles is called an external tangent. Note: Two circles con intersect and have an external tangent (b) Prove: E divides the segment between the centers internally in the ratio of the two radi Internal tangent (c) Prove: E divides the segment between the centers externally in the ratio of the two radil. External tangent ol 4. Below you will find a series of constructions. Prove the statements (al. bl. & (d) and answered. Construction X: Construct an internal tangent to two given circles. Step 1: Divide the line between the centers of the internally in the ratio of the two radil of the circles. Step 2. Locate the midpoints M. of the segmento, and Mof segment 0, Step 3. Draw the circles with centers M, and Mwith respective radii M. and ME From the previous proof TS is an internal tangent to the two circles.

Answer 1

(a). Assume the construction as it is. By the construction, BD MAC. Therefore LB DP = LACP and LDBP = LEAP, since BD is parallel to Ac and the line segments AB 4 CP intersects BD 4 AC. Thus we get for the triangles AAPC 4 A B PD, LACP = LB DP, LAPC = LBP D, 2 il, corresponding angles are LCAP = LDBP. J. equal similar. Hence triangles 4A PC and ABPP are We know that : Similar triangles have proportional sides, that is corresponding sites of trianglesaApe and ABPD opposite to the angles ß and 8 are proportional. This AP = AC = a proved the BY BD

(b). From the figure- LO, ET = LOESI, because they are vertically врrоtе. 8T, 110₂s, and 80 intersect them, so LT,O, E = LS, O₂ E. Consequently, LOIT, E = LORSIEL: Two angles are same, of so the third one has to be same as well] This we get, 40,T, E and 4 O₂ SIE are similar. By similar ang argument as in a We get - O, E = 0, 1 = 2 (proved) Its follows immediately from @ bleause & 4ES, O₂ are similar. M AETO