In the circle O, AB is the diameter, the chord CD intersects the circle O at points c, D, and is parallel to ab. the distance between AB and CD is 5, CD = 24, and ab is calculated

In the circle O, AB is the diameter, the chord CD intersects the circle O at points c, D, and is parallel to ab. the distance between AB and CD is 5, CD = 24, and ab is calculated

For OE ⊥ CD in E and OC, then CE = ed = CD / 2 = 12, OE = 5,
So OC = √ (OE ^ 2 + CE ^ 2) = 13,
So AB = 2oC = 26

As shown in the figure, points c and D are two points on line ab. CD: ab = 2:5, AD + BC = 14cm, find the length of line CD and ab
Picture: --------------
A C D B
Please answer quickly! I will give you a reward

AD+BC=CD+AB=14cm
14/(2+5)=2
CD=2*2=4
AB=2*5=10

Given the line segment AB, extend AB to C to make BC = 12ab, and then extend AB to D reversely to make ad = 23ab. If CD = 26cm, find the length of line segment ab

Let AB = 6x, then BC = 3x, ad = 4x, ∵ AD + AB + BC = DC, ∵ 4x + 6x + 3x = 26, x = 2, ∵ AB = 12

There are two points c and D on the line AB, AC: BC = 5:7, ad: BD = 5:11, and CD = 15cm

AC:BC=5:7
AC:(AC+BC)=5:(5+7)
AC:AB=5:12
AC=5AB/12
AD:BD=5:11
AD:(AD+BD)=5:(5+11)
AD:AB=5:16
AD=5AB/16
CD=AC-AD=5AB/12-5AB/16
=5AB(4-3)/48
=5AB/48=15
AB/48=3
AB=48*3=144cm

As shown in the figure: in the pentagonal ABCDE, ∠ ABC = ∠ AED = 90 °, ∠ BAC = ∠ ead, M is the midpoint of CD, try to judge the size relationship between BM and EM and explain the reason

BM = em. the reasons are as follows: take the midpoint F and g of AC and ad respectively, connect BF, FM, GM and Ge, ∫ ABC = ∠ AED = 90 °, ∫ BF = FA = 12ac, eg = GA = 12ad, ∫ BAF = ∠ ABF, ∫ gae = ∠ GEA, ∫ BFC = 2 ‰ BAC, ∫ EGD = 2 ‰ ead, and ∫ BAC = ∠ ead, ∫ BFC = ∠ EGD, and ∫ M

If the side length of △ ABC is 2, P and Q move on AB and AC respectively, and the segment PQ bisects the area of △ ABC, the value range of segment PQ can be obtained

Let AP = A and AQ = B
Because s △ Apq = 1 / 2S △ ABC
So, 1 / 2 * AB * Sina = 1 / 2 * 1 / 2 * 2 * 2 * Sina
So AB = 2
And because of 0

In a triangle, if two points P Q are taken on the edge AB AC, the line segment PQ bisects the area of △ ABC, and the minimum length of PQ is obtained

Conclusion: if AB and AC are not less than √ K, and AP = AQ = √ K, PQ has a minimum value √ [(1-coa α) &; 2K] if one side of AB and AC is less than √ K, PQ has a minimum value when PQ is the midline of the longer side of AB and ac. if the longer side AB = x, the shorter side AC = y, PQ = √

It is known that in triangle ABC, ab = 5, AC = 4, BC = 3, PQ / / ab

The area of triangle ABC is 4 * 3 / 2 = 6
The area of triangle PQC is: 3
Let PC side length be x, then CQ side length is 3x / 4
Then: 3 = x * (3x / 4) / 2
Then x = double root two

As shown in the figure, in △ ABC, ∠ C = 90 °, ab = 5, BC = 3, P on AC (not coincident with AC), Q on BC, PQ / / AB, if C △ PQC = C quadrilateral pabq, find CP?

Let CP = x, CQ = y ∵ C = 90 °, ab = 5, BC = 3 ∵ AC = 4 ∵ △ the perimeter of PQC = the perimeter of quadrilateral pabq ∵ CP + PQ + CQ = PA + AB + BQ + pqcp + CQ = AP + AB + BQ = ac-cp + AB + bc-cq, that is, x + y = 4-x + 5 + 3-y2x + 2Y = 12 ∵ x + y = 6 (1) ∵ PQ ∥ ab ∥ PQC ∥ ABC ∥ CP / AC = CQ / BC, i.e. X / 4 = Y /

In △ ABC, ab = 5, BC = 3, AC = 4, the point P is on AC, and PQ ‖ AB intersects BC at the point Q (1). When the area of △ PQC and the quadrilateral pabq is equal, the length of PQ is obtained
(2) When the perimeter of △ PQC is equal to that of the quadrilateral pabq, the length of PQ is obtained

（1）5√2/2（2）30/7